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Two positive real numbers have geometric mean $$\sqrt{3}$$ and harmonic mean $$\frac{3}{2}$$ Enter the two numbers, separated by commas.

Aug 6, 2019

$$\sqrt{x y} = \sqrt{3}\\ \dfrac{2}{\dfrac 1 x + \dfrac 1 y}=\dfrac 3 2\\ \dfrac{2xy}{x+y}=\dfrac 3 2\\ \dfrac{6}{x+y}=\dfrac 3 2\\ x+y = 4\\ xy=3\\ x+\dfrac 3 x = 4\\ x^2-4x+3=0\\ (x-3)(x-1)=0\\ x=3,1\\ (x,y) = (1,3), (3,1)$$